%I #22 Aug 16 2018 23:57:15
%S 1,1,1,0,2,1,0,0,1,0,0,0,0,1,2,0,0,1,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,
%T 0,0,0,0,0,0,2,1,0,0,2,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
%U 0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0
%N a(n) = A000699(n) (mod 3).
%C This is a 3-automatic sequence.
%H Muniru A Asiru, <a href="/A304685/b304685.txt">Table of n, a(n) for n = 1..2000</a>
%H John M. Campbell, <a href="/A304685/a304685.pdf">A DFA for enumerating even-order irreducible diagrams modulo 3</a>
%H <a href="/index/Ar#3-automatic">Index entries for 3-automatic sequences</a>
%F For a natural number n, we have that a(n) = 1 if the ternary expansion of n is of the form 100...0 or is of the form 11...1200...0 for an even number of ones in this latter case, allowing runs of integers to be of length 0; a(n) = 2 if the ternary expansion of n is of the form 11...1200...0 if the length of the run of ones is odd; otherwise, a(n) = 0.
%e We have that A000699(5) = 248, with 248 == 2 (mod 3), and the ternary expansion of 5 is given by the tuple (1, 2), so according to the above formula we have that a(5) = 2.
%p a:=proc(n) option remember; if n<=1 then 1; else
%p add((2*i-1)*procname(i)*procname(n-i),i=1..n-1) mod 3; end if; end proc:
%p seq(a(n), n=1..90); # _Muniru A Asiru_, Aug 15 2018
%o (PARI) a(n) = {my(A); A = O(x) ; for( i=1, n, A = x + A * (2 * x * A' - A)); polcoeff(A, n) % 3}; \\ _Michel Marcus_, Jul 04 2018; after A000699
%o (GAP) a:=[1];; for n in [2..90] do a[n]:=(n-1)*Sum([1..n-1],i->a[i]*a[n-i]) mod 3; od; a; # _Muniru A Asiru_, Aug 15 2018
%Y Cf. A000699.
%K nonn
%O 1,5
%A _John M. Campbell_, May 16 2018